Question

If a `11 kg` object moving at `15 ms^-1` slows down to a halt after moving `225 m,` what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

The coefficient of friction is related to the force acting on the object to accelerate (decelerate) it and the normal force: `mu =F/F_N = 5.5/107.8=0.05`.

Explanation:

First we need to calculate the acceleration (deceleration) of the object:

`v^2 = u^2 + 2 ad`

`0^2 = 15^2 -2*a*225`

`a = 0.5 ms^-2`

Now we have an acceleration and a mass and want to find a force... hmm, that sounds a lot like Newton's Second Law to me!

`F=ma=11*0.5=5.5 N`

This is the actual magnitude of the frictional force. The coefficient of friction is defined as:

`mu =F/F_N` where `mu` is the coefficient and `F_N` is the normal force acting. In this case that is the weight force of the object, given by `F_N = mg = 11*9.8 = 107.8 N`

Then `mu =F/F_N = 5.5/107.8=0.05`

Note that a frictional coefficient is a 'dimensionless' number which does not have units (because it is calculated as a force in `N` divided by another force in `N`, so the units cancel).